So I'm writing a program in Python to get the GCD of any amount of numbers.
def GCD(numbers):
if numbers[-1] == 0:
return numbers[0]
# i'm stuck here, this is wrong
for i in range(len(numbers)-1):
print GCD([numbers[i+1], numbers[i] % numbers[i+1]])
print GCD(30, 40, 36)
The function takes a list of numbers. This should print 2. However, I don't understand how to use the the algorithm recursively so it can handle multiple numbers. Can someone explain?
updated, still not working:
def GCD(numbers):
if numbers[-1] == 0:
return numbers[0]
gcd = 0
for i in range(len(numbers)):
gcd = GCD([numbers[i+1], numbers[i] % numbers[i+1]])
gcdtemp = GCD([gcd, numbers[i+2]])
gcd = gcdtemp
return gcd
Ok, solved it
def GCD(a, b):
if b == 0:
return a
else:
return GCD(b, a % b)
and then use reduce, like
reduce(GCD, (30, 40, 36))
Since GCD is associative, GCD(a,b,c,d)
is the same as GCD(GCD(GCD(a,b),c),d)
. In this case, Python's reduce
function would be a good candidate for reducing the cases for which len(numbers) > 2
to a simple 2-number comparison. The code would look something like this:
if len(numbers) > 2:
return reduce(lambda x,y: GCD([x,y]), numbers)
Reduce applies the given function to each element in the list, so that something like
gcd = reduce(lambda x,y:GCD([x,y]),[a,b,c,d])
is the same as doing
gcd = GCD(a,b)
gcd = GCD(gcd,c)
gcd = GCD(gcd,d)
Now the only thing left is to code for when len(numbers) <= 2
. Passing only two arguments to GCD
in reduce
ensures that your function recurses at most once (since len(numbers) > 2
only in the original call), which has the additional benefit of never overflowing the stack.